Differential Equations and Computational Simulations III
J. Graef, R. Shivaji, B. Soni J. & Zhu (Editors)
Electronic Journal of Differential Equations, Conference 01, 1997, pp. 23–39.
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
ftp 147.26.103.110 or 129.120.3.113 (login: ftp)
Uniqueness for a Boundary Identification
Problem in Thermal Imaging ∗
Kurt Bryan & Lester F. Caudill, Jr.
Abstract
An inverse problem for an initial-boundary value problem is considered. The goal is to determine an unknown portion of the boundary of a
region in Rn from measurements of Cauchy data on a known portion of
the boundary. The dynamics in the interior of the region are governed by
a differential operator of parabolic type. Utilizing a unique continuation
result for evolution operators, along with the method of eigenfunction
expansions, it is shown that uniqueness holds for a large and physically
reasonable class of Cauchy data pairs.
1
Introduction
The goal of non-destructive evaluation is to gather information about the interior or other inaccessible portion of some material object from exterior measurements. Thermal imaging is one approach to this problem; a prescribed heat
flux is applied to a portion of the surface of the object and the resulting surface temperature response is measured. From this information one attempts to
determine the internal thermal properties of the object, or the shape of some
unknown, inaccessible portion of the boundary. Thermal imaging holds promise
as a tool for corrosion detection in aircraft, and has found utility in industrial
applications. The interested reader is referred to [2], and the references therein,
for a discussion of some of these applications. Thermal imaging methods have
also found application in the medical field. For example, infrared thermography
has been used to investigate the distribution and structure of skin blood vessels;
this has implications regarding potential recovery from burn injuries ([3]), and
also bears on the selection of donor sites for skin grafts ([5]).
We are interested in the use of thermal imaging for the detection of so-called
“back surface” corrosion and damage. The most elementary model of such a
process is simple material loss which leads to a change in the surface profile of
the object’s boundary. This is the model we have chosen for this paper. Our
∗ 1991 Mathematics Subject Classifications: 35A40, 35J25, 35R30.
Key words and phrases: Inverse problems, non-destructive testing, thermal imaging.
c
1998
Southwest Texas State University and University of North Texas.
Published November 12, 1998.
Partially supported by NSF Grant DMS-9623279.
23
24
Uniqueness for a Boundary Identification Problem
long-term goal is to develop a reliable method for determining the presence and
extent of corrosion. Of course, any such method requires a sound theoretical
foundation. To this end, our focus in this work is on the issue of uniqueness—
under what conditions do the proposed data measurements provide sufficient
information from which to determine the shape of an unknown portion of the
object’s boundary? This problem may be formulated mathematically as an
inverse problem for the heat equation. More precisely, let Ω ⊆ Rn represent
the object to be imaged. We assume that the surface ∂Ω of Ω is piecewise C 2 .
We use Γ to denote the “known,” accessible portion of ∂Ω, and we assume that
both Γ and ∂Ω \ Γ have nonzero surface measure as subsets of ∂Ω. Let S0
denote some open portion of Γ with positive measure and let the applied heat
flux g(t, x) be defined for each (t, x) ∈ R+ ; ∂Ω with support in S0 . With some
rescaling we model the propagation of heat through Ω with an initial-boundary
value problem for the heat equation,
ut (t, x) − ∆x u(t, x) =
∂u
(t, x) =
∂η
u(0, x) =
0,
for t ∈ R+ , x ∈ Ω ,
g(t, x) ,
u0 (x) ,
for t ∈ R+ , x ∈ ∂Ω ,
for x ∈ Ω.
(1)
(2)
(3)
Here u(t, x) denotes the temperature in the domain Ω at the point x at time t, ut
the derivative of u with respect to t, and η an outward unit normal vector field
on ∂Ω. Throughout this paper, we will refer to (1)-(3) collectively as (IBVP).
Let S1 ⊂ Γ denote the portion of the boundary on which we take temperature
measurements. We consider the following inverse problem: Does knowledge of
u(t, x) on S1 for some time period t0 < t < t1 uniquely determine ∂Ω \ Γ?
Specifically, suppose Ω1 ⊆ Rn and Ω2 ⊆ Rn with Γ contained in ∂Ω1 ∩ ∂Ω2 .
For j = 1, 2, let uj (t, x) be the solution of (1)-(3) with Ω replaced by Ωj . If,
u1 = u2 on (t0 , t1 ) × S1 , must it be true that ∂Ω1 \ Γ = ∂Ω2 \ Γ?
Remark. Implicit in the formulation of (IBVP) is the assumption that on the
unknown part of the boundary the condition ∂u
∂η = 0 holds, so that the back
surface acts as a perfect insulator. This is only a first approximation in most
situations. In Section 5 we discuss other boundary conditions in which the back
surface loses heat to the ambient environment.
The answer to the uniqueness problem posed in the present paper will be
seen to depend on certain properties of the domain Ω, the initial condition u0 ,
and the flux g. We will show that uniqueness holds for constant u0 and any
non-zero flux g. For non-constant u0 one can impose reasonable conditions on
the flux g to ensure uniqueness, provided Ω is bounded. The case in which the
flux g is time-periodic was analyzed in [2].
This paper is organized as follows. The case of constant initial condition and
non-constant flux is analyzed in §2, where uniqueness is proved. In §3 we derive
a useful eigenfunction representation and associated estimates for solutions of
(IBVP), which are used in §4 to prove a uniqueness result for bounded domains.
Kurt Bryan & Lester F. Caudill, Jr.
25
In §5, we extend our results to include other possibilities for the boundary
conditions on Γ.
The fact that uniqueness for the inverse problem fails without additional
hypotheses on the ingredients in (1)-(3) may be illustrated by a simple example
in R2 . Let Ω1 be the rectangle defined by 0 < x < 2π, 0 < y < π. Let Ω2 be
Ω1 minus the rectangle 23 π < x < 43 π, 0 < y < 23 π, so that Ω1 and Ω2 share the
“known” top boundary Γ = {(x, y) : 0 < x < 2π, y = π}. Let u(t, x, y) be the
function
3
3
9
u(t, x, y) = e− 2 t cos( x) cos( y) ,
2
2
and set u1 ≡ uΩ1 and u2 ≡ uΩ2 . One may verify directly that, for j = 1, 2,
∂uj
∂η
9
∂u
= e− 2 t cos( 32 x) on Γ while ∂ηj = 0 everywhere else on ∂Ωj . Both u1 and u2
satisfy (1) with the same initial condition, with the same Cauchy data on Γ, and
so in this case uniqueness fails. Analogous counter-examples can be constructed
in other dimensions.
2
Constant Initial Condition
In this section we develop a uniqueness result for the case in which the initial
condition u0 (x) is constant. The only condition on the applied flux g(t, x) is
that it be regular enough for (IBVP) to possess a unique solution, e.g., g ∈
C(R; L2 (∂Ω)).
In what follows, we will require the following lemma. Let Ω1 and Ω2 represent two objects which share the “known” boundary portion Γ. One can see
that there exists a connected component of Ω1 ∩ Ω2 which has Γ as part of its
boundary. We shall denote this component by Ω0 .
Lemma 2.1 Let (u1 , Ω1 ) and (u2 , Ω2 ) each satisfy (1)-(3). If u1 = u2 on
(0, T ) × S1 for some time T > 0, then u1 = u2 on (0, T ) × Ω0 .
In proving this lemma, we will make use of the following unique continuation
result for parabolic equations. Its proof is based on the derivation of inequalities
of Carleman type, and is omitted here. The interested reader is referred to the
work of Saut and Scheurer [8].
Lemma 2.2 Let Ω be a connected
open set in Rn and Q = (−T, T ) × Ω. Let
2
2
u ∈ L (−T, T ); Hloc (Ω) be a solution of ut − ∆u = 0 which vanishes in some
open subset O of Q. Then u vanishes in the horizontal component of O.
Note: Following Nirenberg [7], we define the horizontal component Oh of O
to be the union of all open hyperplanes of the form t = constant in Q which
have nonempty intersection with O.
26
Uniqueness for a Boundary Identification Problem
Proof of Lemma 2.1. The function w ≡ u1 − u2 obeys
wt − ∆w
∂w
w=
∂η
w(x, 0)
=
0,
on (0, T ) × Ω0 ,
=
0,
on (0, T ) × S1 ,
=
0,
on Ω0 .
We can choose some open connected subset I with I¯ ⊂ S1 and open ball
B ⊂ Rn such that B ∩ ∂Ω0 = I. Let ΩB = B \ Ω0 set Ω̃ ≡ Ω0 ∪ ΩB . Define the
function

 w, (0, T ) × Ω0 ;
w̃ ≡ 0, (0, T ) × ΩB ;

0, (−T, 0] × Ω̃.
For a smooth test function φ,
Z TZ
w̃ [φt + ∆φ] dxdt = 0 ,
0
Ω̃
so that w̃ satisfies (1) on Ω̃. Using standard
parabolic
arguments (see,
regularity
e.g., [6]), one can show that w̃ ∈ H 1 (−T, T ); H 2 Ω̃ .
Make the identifications Q ≡ (−T, T )× Ω̃ and O ≡ (−T, T )×int(ΩB ) (in this
case, the horizontal component of O is Q) to see that w̃ satisfies the hypotheses
of Lemma 2.2. We conclude that w̃ vanishes on Q and so u1 = u2 on (0, T ) × Ω0 .
We now present the main result of this section.
Theorem 2.1 Let (u1 , Ω1 ) and (u2 , Ω2 ) be solutions of (1)-(3), with (S0 ∪S1 ) ⊆
(∂Ω1 ∩ ∂Ω2 ). Suppose u0 (x) = u0 , a constant, and suppose that there is some
time T > 0 for which the applied flux g(t, x) is not identically zero on (0, T )×S0.
If u1 = u2 on (0, T ) × S1 then Ω1 = Ω2 and u1 = u2 on Ω1 = Ω2 .
proof By replacing uj with uj − u0 , for j = 1, 2, if necessary, it suffices to
consider the case u0 = 0. Suppose that Ω1 6= Ω2 , and let Ω0 be as above. Then
there exists some nonempty connected component D, sharing a portion of its
boundary with ∂Ω0 , of either Ω1 \ Ω2 or Ω2 \ Ω1 . Let us suppose the latter, so
that u2 is defined and satisfies (1) on D. The boundary ∂D of D is comprised of
a portion Γ1 of ∂Ω1 and Γ2 of ∂Ω2 . On Γ2 we know that the normal derivative of
u2 is identically zero; on Γ1 , we know that the normal derivative (from inside Ω1 )
of u1 is zero, and since u2 ≡ u1 on Ω0 (by Lemma 2.1) and u2 is smooth across
Γ1 , we conclude that the normal derivative of u2 vanishes on the boundary of
D. Since u2 satisfies equation (1) with zero initial data on D, this forces u2 ≡ 0
on (0, T ) × D. Finally, by extending u2 to be zero on (−T, 0] × (Ω0 ∪ D), we
may appeal to Lemma 2.2 to conclude that u2 ≡ 0 on (0, T ) × (Ω0 ∪ D). This
in turn implies that the flux g is identically zero on (0, T ) × S0 , a contradiction,
and we must conclude that Ω1 = Ω2 , as asserted.
Kurt Bryan & Lester F. Caudill, Jr.
3
27
Eigenfunction Expansion
In this section we record a useful eigenfunction expansion for the function u(t, x)
which satisfies the parabolic initial-boundary value problem (IBVP) (1)-(3).
The technique, as well as the derivation of the accompanying estimates, are
standard, and we have spared the reader the details. Instead, we defer to [9], or
virtually any text on classical PDE.
We assume that the initial condition u0 belongs to L2 (Ω) and that for all t >
0 the applied flux g(t, x) belongs to C 1 ((0, T ); L2 (∂Ω)), the space of continuously
differentiable functions from (0, T ) to L2 (∂Ω). We seek a solution u(t, x) to
(IBVP) in the space C((0, T ); L2 (Ω)); for such a solution the derivatives of u
with respect to t and x are not well-defined, and so we cast (IBVP) into a weak
form. Multiply equation (1) by a smooth test function φ(t, x) with φ(T, x) ≡ 0
and ∂φ
∂η = 0 on ∂Ω, and then integrate over (0, T ) × Ω. Integrate the term
involving φut by parts in t use Green’s second identity on the term involving
φ 4 u to obtain
Z
Z TZ
Z TZ
u0 (x)φ(0, x) dx +
u (φt + ∆φ) dx dt +
φg dSx dt = 0. (4)
Ω
Ω
0
0
∂Ω
The restriction of the L2 (Ω) function u to ∂Ω is not well-defined, but since
∂φ
∂φ
∂η = 0 on ∂Ω the boundary integral involving u ∂η vanishes. This is our weak
form of (1) - (3).
In preparation for the eigenfunction expansion, let {λk , ψk (x)}, k = 0, 1, . . .
be an eigensystem for −∆ on Ω with homogeneous Neumann boundary conditions, so that
∆ψk + λk ψk
∂ψk
∂η
=
0 in Ω,
=
0 on ∂Ω..
The eigenvalues λk are non-negative; order them by magnitude, so λk ≤ λk+1 .
k
With the boundary condition ∂ψ
∂η = 0, the first eigenvalue λ0 = 0, is simple, and has a constant eigenfunction. We normalize the eigenfunctions so that
kψk kL2 (Ω) = 1 for all k, and so obtain an orthonormal basis for L2 (Ω). The funcp
tion ψ0 (x) is constant and ψ0 (x) = 1/ |Ω|. Orthogonality of the eigenfunctions
then implies that
Z
ψk (x) dx = 0 , k ≥ 1 .
Ω
In later sections we will make use of the following standard estimate for
solutions to Poisson’s equation with Neumann boundary conditions.
Lemma 3.1 Let f1 ∈ L2 (Ω), f2 ∈ L2 (∂Ω), and let ψ(x) ∈ H 1 (Ω) satisfy
4ψ
∂ψ
∂η
Z
ψ(x) dx
Ω
= f1 in Ω,
= f2 on ∂Ω,
= 0,
28
Uniqueness for a Boundary Identification Problem
Then
kψkH 1 (Ω) ≤ C(kf2 kL2 (∂Ω) + kf1 kL2 (Ω) )
where C depends on the domain Ω.
The main result of this section is
Lemma 3.2 The solution u(t, x) to (4) is unique in C(R+ ; L2 (Ω)), and can be
expanded as
d0
1
u(t, x) = v(t, x) + p
+
|Ω| |Ω|
Z
t
G(s) ds +
0
∞
X
Tk (t)ψk (x)
(5)
k=1
where v(t, x) defined on R+ × Ω denotes the unique function which satisfies the
family of elliptic problems (indexed by t)
4x v
=
∂v
∂η
Z
1
G(t) in Ω,
|Ω|
(6)
= g(t, x) on ∂Ω,
v(t, x) dx
= 0,
Ω
and
Z
ck (t)
g(t, x) dSx ,
Z∂Ω
=
(u0 (x) − v(0, x))ψk (x) dx,
Ω
Z
= −
vt ψk (x) dx , k > 0 ,
c0 (t)
=
G(t)
dk
Tk (t)
(7)
=
(8)
(9)
Ω
G(t)
p ,
|Ω|
= dk e
−λk t
Z
+
0
t
ck (s)e−λk (t−s) ds, k > 0,
(10)
where |Ω| denotes the measure of Ω, dSx denotes surface measure on ∂Ω, and
vt is the derivative of v(t, x) with respect to t (which exists with the given hypotheses). We also have the estimate
∞
X
Tk2 (t)
≤C
e
−2λ1 t
ku0 kL2 (Ω) +
k=1
where C is a constant which depends on Ω.
tkgt (t, ·)k2L2 (∂Ω)
2λ1
!
(11)
Kurt Bryan & Lester F. Caudill, Jr.
4
29
Uniqueness for Bounded Regions
We now consider the more general case in which the initial condition u0 need
not be constant. Here we will assume that Ω is a bounded region. The essential
idea in this section is simple. We note from the proof of Theorem 2.1 that if
uniqueness fails then there must be some “insulated” region D inside Ω. Within
such a region, heat neither enters nor leaves, so that the average temperature
of D cannot increase with time. This is the basis of the argument that follows:
intuitively, if the applied flux g pumps enough heat into Ω over a long enough
period then no region D can remain at the same average temperature, and so a
uniqueness result must hold. We make this physical argument precise below.
Theorem 4.1 Let g(t, x) denote a flux in the class C 1 (R; L2 (∂Ω)) supported
for x ∈ S0 with kg(t, ·)kL2 (∂Ω) ≤ M0 for all t > 0 and kgt (t, ·)kL2 (∂Ω) ≤ M1 for
all t > 0. Suppose also that G(t) defined by equation (7) satisfies G(t) ≥ G0 > 0
for all t. Let u(t, x) be the solution to (1)-(3) or its weak form (4) (the initial
condition u0 is not considered known). Then knowledge of u(t, x) for 0 < t < ∞
and x ∈ S1 uniquely determines the region Ω and the initial condition u0 .
Proof of Theorem 4.1. Suppose that u1 and u2 are solutions to the weak
form (4) of (IBVP) on domains Ω1 and Ω2 , respectively, with initial conditions
u1 (0, x) = u0 (x) and u2 (0, x) = ũ0 (x). We assume that temperature measurements are taken on an open subset S1 ⊂ (∂Ω1 ∩ ∂Ω2 ) and the same flux g(t, x)
applied on an open subset S0 ⊂ (∂Ω1 ∩ ∂Ω2 ). We will show that there is some
time T > 0 such that measurements of u1 and u2 on (0, T ) × S1 must differ.
We proceed by contradiction. Assume that u1 ≡ u2 on (0, ∞) × S1 , and as
before, let Ω0 denote the connected component of Ω1 ∩ Ω2 for which Γ ⊆ ∂Ω0 .
Let w = u1 − u2 . We will show that w(t, x) ≡ 0 on (0, ∞) × Ω0 . To see this note
that the function w satisfies
∂w
− 4w = 0 , in Ω0 × (0, ∞) ,
∂t
(12)
with w = ∂w
∂η = 0 on S1 × (0, ∞). Let p be a point in S1 and B a ball centered
at p such that B ∩ ∂Ω0 ⊂ S1 . Let B0 denote that portion of B which lies outside
Ω0 . Define w̃(t, x) on Ω0 ∪ B0 as
w(t, x), x ∈ Ω0
w̃(t, x) =
0
x ∈ B0
Standard regularity results (see [6]) show that w ∈ L2 ((0, T ); H 2 (Ω0 )) for any
2
2
0
T > 0. Since w = ∂w
∂η ≡ 0 on S1 , it is easy to check that w̃ ∈ L ((0, T ); H (Ω ∪
B0 )). The function w̃ vanishes on B0 × (0, ∞) and we conclude from Lemma
2.2 (with the minor alteration −T → 0) that w̃ vanishes on Ω0 × (0, ∞). This
shows that u1 ≡ u2 on Ω0 × (0, ∞). Also, since (12) has a unique solution for
given initial and boundary conditions, we conclude that u0 = ũ0 on Ω0 .
If Ω1 6= Ω2 then either Ω1 \ Ω2 or Ω2 \ Ω1 contains a nonempty connected
component D for which ∂D ∩ ∂Ω0 has positive surface measure. For specificity,
30
Uniqueness for a Boundary Identification Problem
we assume that D ⊂ (Ω2 \ Ω1 ). The boundary of D consists of portions of
∂Ω1 \ (S0 ∪ S1 ) and ∂Ω2 \ (S0 ∪ S1 ). On these portions of the boundary the
applied flux g is identically zero. Standard regularity results then show that u2
is a classical solution to the heat equation and smooth on D̄, and we therefore
2
have ∂u
∂η ≡ 0 on ∂D. Since D is bounded and u2 is smooth,
Z
Z
Z
Z
∂u2
∂u2
d
dx =
dSx = 0.
u2 (t, x) dx =
4u2 dx =
dt D
∂t
D
D
∂D ∂η
R
The integral D u2 (t, x) dx on the left is just the total thermal energy inside
D. We have thus shown that the assumption that u1 = u2 on (0,
R ∞) × S1 and
Ω1 6= Ω2 force the existence of an insulated region D for which D u2 (t, x) dx is
constant. We will show that this is impossible for an applied flux g(t, x) of the
form specified in the statement of the theorem.
Let u2 (t, x) be expressed via an eigenfunction expansion as in equation (5).
Integrating over D shows that
Z
Z
Z
Z X
∞
d0 |D|
|D| t
u2 (t, x) dx =
v(t, x) dx + p
G(s) ds +
Tk (t)ψk (x) dx
+
|Ω2 | |Ω2 | 0
D
D
D k=1
(13)
where Tk (t) is defined by equation (10), d0 by equation (8), and v satisfies (6)
Rt
with Ω replaced by Ω2 . Since G(t) ≥ G0 for all t, the integral 0 G(s) ds grows
at least as fast as G0 t; however, the other terms
R in the equation can be shown
to be o(t) as t → ∞, and this will show that D u2 (t, x) dx cannot be constant.
The first integral on the right side of equation (13) is bounded in t, which
can be proved by noting that
Z
p
v(t, x) dx ≤
|D|kv(t, ·)kL2 (D)
D
p
≤
|D|kv(t, ·)kL2 (Ω) ,
and applying Lemma 3.1 with the fact that kg(t, ·)kL2 (∂Ω) ≤ M0 .
The second term in equation (13) is constant and, therefore, bounded in t.
The last term can be estimated by noting that
∞
Z ∞
X
X
p
Tk (t)ψk (x) dx ≤
|D| Tk (t)ψk (x) dx
,
D
k=1
k=1
L2 (D)
∞
X
p
|D| Tk (t)ψk (x) dx
,
≤
k=1
L2 (Ω)
v
u∞
p uX
|D|t
Tk2 (t).
(14)
=
k=1
Combining (14) with the estimate (11) in Lemma 3.2 shows that
1/2
Z X
∞
p
M1 t
−2λ1 t
2
Tk (t)ψk (x) dx ≤ C |D| e
ku0 kL2 (Ω) +
.
2λ1
D
k=1
(15)
Kurt Bryan & Lester F. Caudill, Jr.
31
The quantity on the right side of (15) is clearly o(t), and so grows more slowly
Rt
than 0R G(s) ds. Equation (13) then shows that for sufficiently large t the integral D u2 (t, x) dx must increase, a contradiction that proves Theorem 4.1.
If in addition to the conditions above g is analytic in t (for example, if g is
independent of t, so g = g(x)) then we can do better. Suppose that g(t, x) ∈
C ω ((0, T ); L2 (∂Ω)), i.e. for each t0 > 0 there is some δ > 0 such that g(t, x) can
be written as
∞
X
(t − t0 )k
gk (x)
g(t, x) =
k!
k=0
for all t with |t − t0 | < δ, where gk ∈ L2 (∂Ω). In this case the solution to (1)-(3)
is analytic in t,
∞
X
(t − t0 )k
u(t, x) =
uk (x)
k!
k=0
where uk ∈ L2 (Ω). Suppose that two domains Ω1 and Ω2 give rise to the same
temperature measurements on (t1 , t2 ) × S1 with t1 < t2 . Arguing as in the proof
of Theorem 4.1 we find that u1 ≡ u2 on (t1 , t2 ) × Ω0 , but since u1 and u2 are
analytic in t we have u1 ≡ u2 on (0, ∞) × Ω0 . The rest of the proof of Theorem
4.1 remains unchanged and we have
Theorem 4.2 Let g(t, x) denote a flux in the class C ω (R; L2 (∂Ω)) supported
for x ∈ S0 with kg(t, ·)kL2 (∂Ω) ≤ M0 for all t > 0 and kgt (t, ·)kL2 (∂Ω) ≤ M1 for
t > 0. Suppose also that G(t) defined by equation (7) satisfies G(t) ≥ G0 > 0
for all t. Let u(t, x) be the solution to the IBVP (1)-(3) or its weak form (4)
(the initial condition u0 is not considered known). Then knowledge of u(t, x)
for any open time interval 0 < t1 < t < t2 and x ∈ S1 uniquely determines the
region Ω and the initial condition u0 .
5
Other Boundary Conditions
The results of the previous section show that we can uniquely identify the unknown portion of the surface of Ω if we pump in enough heat for a long enough
∂u
= 0 on
time. However, in this situation the insulated boundary condition ∂n
∂Ω becomes physically less realistic. Those portions of the boundary on which
a nonzero flux is not applied will tend to lose heat to the surrounding environment. In this section we consider uniqueness results under boundary conditions
which model this heat loss. The proofs are quite similar to those of the previous
section.
Suppose that u(t, x) satisfies the initial-boundary value problem
∂u
− 4u =
∂t
∂u
+ αu =
∂n
u(0, x) =
0 on R+ × Ω
(16)
g(t, x), on R+ × ∂Ω
(17)
u0 (x) on Ω
(18)
32
Uniqueness for a Boundary Identification Problem
with α > 0 and g supported for x ∈ S0 ⊂ ∂Ω. The Robin boundary condition
∂u
∂n + αu = 0 corresponds to a Newton-cooling type of heat loss on the boundary
with ambient temperature scaled to zero; note that we have assumed that the
loss term −αu applies even on S0 , where the flux g is applied.
The solution u to the initial-boundary value problem (16)-(18) can be represented with an eigenfunction expansion, as
u(t, x) = v(t, x) +
∞
X
Tk (t)ψk (x)
(19)
k=0
where v(t, x) satisfies the family of elliptic problems (indexed by t)
4x v
=
0 in Ω,
(20)
∂v
+ αv
∂n
=
g on ∂Ω,
(21)
Tk (t) is defined by
Tk (t) = dk e
and
ck (t)
dk
−λk t
+e
−λk t
Z
0
t
ck (s)eλk s ds,
Z
= −
vt ψk (x) dx,
Ω
Z
=
(u0 (x) − v(0, x))ψk (x) dx
(22)
(23)
Ω
and finally, {λk , ψk (x)} is an orthonormal eigensystem for −4 with the Robin
boundary conditions, so for each k,
4ψk + λk ψk
∂ψk
+ αψk
∂n
=
0 in Ω,
=
0, on ∂Ω .
We order the eigenvalues by magnitude. It is easy to check that all eigenvalues
are strictly positive.
The next result gives sufficient conditions on the induced flux g which guarantee uniqueness. As before, we assume that we have measurements of u(t, x)
for x ∈ S1 ⊂ ∂Ω. Loosely speaking, we require the flux to be nonnegative and
decaying in time, but not too quickly. More precisely, we have
Theorem 5.1 Let (u1 , Ω1 ) and (u2 , Ω2 ) be solutions of (16)-(18) with (S0 ∪
S1 ) ⊆ (∂Ω1 ∩ ∂Ω2 ). Suppose that the applied flux g(t, x) ∈ C 1 (R; L2 (∂Ω)) and
is supported in S0 for each t. Also, assume that
1. g(t, x) is not identically zero.
2. g(t, x),
∂g
∂t (t, x)
≥ 0 for all x and t.
Kurt Bryan & Lester F. Caudill, Jr.
33
3. kgt (t, ·)kL2 (∂Ω) → 0 as t → ∞ in such a way that
sups>t kgt (s, ·)k2L2 (∂Ω) = o ln1 t as t → ∞.
Then u1 ≡ u2 on R+ × S1 implies that Ω1 = Ω2 .
In the case where Ω belongs to R2 or R3 , one may relax the decay condition 3.
in this result. More precisely,
Theorem 5.2 In space dimension 2 or 3, Theorem 5.1 holds, with hypothesis
3. relaxed to
30 . kgt (t, ·)kL2 (∂Ω) → 0 as t → ∞.
We shall prove Theorem 5.1 first. Afterward, we will indicate the changes
necessary to establish Theorem 5.2.
Proof of Theorem 5.1. Our proof proceeds by contradiction. Suppose Ω1 6=
Ω2 . The same reasoning as in the proof of Theorem 2.1 shows that we must
have some nonempty region D ⊂ Ω (where Ω is Ω1 or Ω2 ), with ∂D ∩∂Ω0 having
positive surface measure, on which
∂u
− 4u = 0 on R+ × D
∂t
∂u
+ αu = 0, on R+ × ∂D
∂n
u(0, x) = u0 (x) on D
(where u is either u1 or u2 ). (This is theR same Ω0 as in the proof of Theorem
4.1.) We first observe that the integral D u(t, x) dx must tend exponentially
rapidly to zero as t → ∞. To see this, note that u(t, x) can be expanded on D
in terms of eigenfunctions
u(t, x) =
∞
X
dk e−λ̃k t ψ̃k (x)
(24)
k=0
Z
with
dk =
D
u0 (x)ψ̃k (x) dx.
and {λ̃k , ψ̃k (x)} is an eigensystem for −4 on D with boundary conditions
∂ ψ̃k
∂n + αψ̃k = 0 on ∂D. Again, the eigenvalues are strictly positive. From the
representation (24)
Z
p
u(t, x) dx ≤
|D|kukL2 (D)
D
=
O(e−λ̃0 t )
where λ̃0 > 0 is the smallest eigenvalue for the above eigensystem.
(25)
34
Uniqueness for a Boundary Identification Problem
We shall complete the proof of Theorem 5.1 by contradicting relation (25)
in the following way: We shall show that, under the hypotheses on g,
Z
Z
lim
u(t, x) dx −→
v(t, x) dx.
(26)
t→∞ D
D
Z
v(t, x) dx is bounded away from zero, uniformly in t.
(27)
D
R
From these two facts, it is clear that D u(t, x) dx must be bounded away from
zero, uniformly in t, the desired contradiction to (25).
To establish (26), we first show that ku − vkL2 (Ω) → 0 as t → ∞. Note that
(19) and (22) imply
ku − vk2L2 (Ω)
=
≤
∞
X
Tk2 (t)
k=0
∞ Z t
X
2
k=0
0
ck (s)e
−λk (t−s)
2
ds
+ o(1)
(28)
where the last equality follows from the fact that λk > 0 for each k. The integral
appearing inside the sum on the right can be bounded as
!2
Z
2
Z
Z
t
0
ck (s)e−λk (t−s) ds
β
=
t
β
ck (s)e−λk (t−s) ds
Z β
e−2λk (t−β) − e−2λk t
c2k (s) ds
λk
0
Z t
1 − e−2λk (t−β)
c2k (s) ds
+
λk
β
≤
0
ck (s)e−λk (t−s) ds +
(29)
where β = β(t) ∈ (0, t) is to be specified in a moment. From the bounds (28)
and (29) we conclude that
Z β X
−2λ0 (t−β)
∞
− e−2λ0 t
e
2
c2k (s) ds
ku − vkL2 (Ω) ≤ C
λ0
0 k=0
!
Z tX
∞
1 − e−2λ0 (t−β)
c2k (s) ds
(30)
+
λ0
β
k=0
for some constant C, where we have interchanged the summation and integral
−λk t
−λk t
−e−2λk t
and 1−eλk
are
for the convergent series and used that fact that e
λk
decreasing functions of λk for λk , t > 0.
P∞
Next, note that from equation (23) we have k=0 ck (t)2 = kvt (t, ·)k2L2 (Ω) ,
where vt satisfies
4vt
∂vt
+ αvt
∂η
=
0 in Ω,
(31)
=
gt on ∂Ω
(32)
Kurt Bryan & Lester F. Caudill, Jr.
35
Standard estimates similar to Lemma 3.1 show that we can bound
kvt kL2 (Ω) ≤ Ckgt kL2 (∂Ω) .
(33)
It then follows from (30) and (33) that
ku −
vk2L2 (Ω)
≤
C ( (e
−2λ0 (t−β)
−2λ0 t
C ( (e
−2λ0 (t−β)
−2λ0 t
Z
β
−e
)
kgt (s, ·)k2L2 (∂Ω) ds
0
Z t
−2λ0 (t−β)
)
kgt (s, ·)k2L2 (∂Ω) ds )
+(1 − e
β
≤
+(1 − e
−e
−2λ0 (t−β)
)β · sup kgt (s, ·)k2L2 (∂Ω)
0<s<β
)(t − β) sup kgt (s, ·)k2L2 (∂Ω) )
β<s<t
(34)
(the 1/λ0 factor has been absorbed into the constant C). For r ≥ 0, define
(r) = sups>r kgt (s, ·)k2L2 (∂Ω) , and observe that (r) has the following properties:
• For each r, (r) ≥ 0 and 0 (r) ≤ 0.
• (0) < ∞ and (r) → 0 as r → ∞.
From equation (34), we have, in terms of (r),
ku − vk2L2 (Ω)
≤ C ( (e−2λ0 (t−β) − e−2λ0 t )β · (0)
+(1 − e−2λ0 (t−β) )(t − β)(β) )
(35)
We now specify, for each fixed t > 0, a corresponding value of β implicitly by
the relation
β + K ln β = t ,
where K ∈ R+ satisfies K > 2λ1 0 . It is simple to check that this relation defines
β uniquely as a function of t, and that β(t) → ∞ as t → ∞. Furthermore,
t − β = K ln β → ∞
as t → ∞. With β(t) so defined, we note that
(t − β)(β) = K ln(β)(β) → 0
(36)
as t (and hence β) → ∞, by virtue of the decay condition on . Also,
βe−2λ0 (t−β) = βe−2λ0 K ln β = β 1−2λ0 K → 0
(37)
as t → ∞, since 2λ0 K > 1. In light of (36) and (37), we see that the right-hand
side of (35) decays to 0 as t → ∞. From this, we conclude that ku − vkL2 (Ω) → 0
as t → ∞. In fact, ku − vkL2 (D) → 0 for any D ⊂ Ω, so that
Z
p
(u − v) dx ≤ ku − vkL1 (D) ≤ |D|ku − vkL2 (D) → 0 ,
D
36
Uniqueness for a Boundary Identification Problem
from which we conclude
Z
Z
D
u(t, x) dx →
v(t, x) dx
(38)
D
as t → ∞, which is (26).
It remains only to establish (27). To this end, let us first consider the case in
which g(t, x) ∈ C 1 (R; C 2 (∂Ω)). Then the function v(t, ·) ∈ C 2 (Ω̄) for all t. We
have by the maximum principle that the minimum value of v(t, x) on Ω̄ occurs
∂v
at a point on ∂Ω at which ∂v
∂η ≤ 0. At such a point, αv = g − ∂η ≥ 0, from
whichR we conclude that v(t, x) ≥ 0 for x ∈ Ω. In particular, for any D ⊂ Ω we
have D v(t, x) dx ≥ 0, with equality if and only if v(t, x) ≡ 0. Since
v ≡ 0 if and
R
only if g ≡ 0 (from (31)-(32)),
the
hypotheses
on
g
imply
that
v(t,
x) dx > 0
D
R
for each t. (In particular, RD v(0, x) dx > 0.) Furthermore, since gt ≥ 0, the
same reasoning shows that D vt (t, x) dx ≥ 0 for any D ⊂ Ω and all t > 0.
Consequently, for each t > 0,
Z
Z
Z
Z t
∂
v(t, x) dx =
v(0, x) dx +
v(s, x) dxds
D
D
0 ∂s D
Z tZ
Z
v(0, x) dx +
vt (s, x) dxds
=
D
0
ZD
v(0, x) dx > 0 ,
≥
D
where we have used the fact that D is bounded and smooth enough to interchange the order of integration and differentiation. This establishes (27) for
g(t, x) ∈ C 1 (R; C 2 (∂Ω)).
Finally, we note that the same conclusion holds if g(t, ·) is merely L2 (∂Ω),
rather than C 2 (∂Ω), for we can approximate any non-negative g ∈ L2 (∂Ω)
arbitrarily closely (in L2 (∂Ω)) with a non-negative function g̃(t, ·) ∈ C 2 (∂Ω).
From the standard estimate
kv − ṽkL2 (Ω) ≤ Ckg − g̃kL2 (∂Ω)
where ṽ satisfies the boundary value problem (20)-(21) with g replaced by g̃, we
conclude that kv − ṽkL2 (D) can be made arbitrarily small. Since
Z
Z
p
v(t, x) dx −
≤ |D|kv − ṽkL2 (D) ≤ Ckg − g̃kL2 (∂Ω) ,
ṽ(t,
x)
dx
D
R
D
R
and since D ṽ(t, x) dx > 0 uniformly in t, we conclude that D v(t, x) dx > 0
uniformly in t also. This establishes (27), and completes the proof.
Proof of Theorem 5.2. Noting that the decay condition o ln1 t was used
only to establish (38), the preceding proof also works for Theorem 5.2, provided
we show that this relation still holds. Let Ω ⊆ Rn with n = 2 or n = 3, and set
2
∞ Z t
X
−λk (t−s)
φ(t) ≡
ck (s)e
ds
.
k=0
0
Kurt Bryan & Lester F. Caudill, Jr.
37
In light of (28), it suffices to show that
lim φ(t) = 0 .
t→∞
To this end, set
M (t) ≡ sup k{ck (s)}kl2 (k) .
t<s<∞
(Recall that k{ck (s)}kl2 (k) is of the same order as kgt (t, ·)kL2 (∂Ω) as t → ∞.)
Note that |ck (s)| ≤ M (t) for s > t and all k. We can estimate
!2
Z t/2
Z t
∞
X
ck (s)e−λk (t−s) ds +
ck (s)e−λk (t−s) ds
φ(t) =
k=0
0
t/2
≤ 2M 2 (0)e−λ0 t/2
∞
X
k=0
In R2 or R3 , we have
P∞
1
k=0 λ2k
∞
X 1
1
+ 2M 2 (t/2)
2
λk
λ2k
< ∞ , so that (39) yields
φ(t) ≤ C1 e−λ0 t/2 + C2 M 2 (t/2) .
By hypothesis 30 ., M 2t → 0 as t → ∞, so that (38) holds.
6
(39)
k=0
Concluding Remarks
We have examined a variety of settings in which the Cauchy data for the heat
equation uniquely determines the shape of the region on which the heat equation is defined. Specifically, if the initial condition is constant over the region
of interest, then the Cauchy data—temperature and heat flux—on any open
portion of the boundary of the region over any time interval determines the
shape of the region. In the more general case in which the initial condition is
not necessarily constant, the Cauchy data for the time interval (0, ∞) uniquely
determines the shape of a bounded region Ω, provided that the data satisfy
certain reasonable conditions. For insulate boundary conditions ∂u
∂η = 0 on the
unknown portion B of the boundary, the prescribed flux g(t, x) on ∂Ω \ B must
provide a net positive flux at all times and be bounded away from zero, and gt
must be bounded. For the Robin boundary condition ∂u
∂η + αu = 0, we require
the flux to be positive at all points and times, and obey a certain decay property
in time.
The techniques employed in this analysis can be used to investigate other
types of boundary conditions. The choices presented here reflect sensible conditions within the context of the particular physical situation—remote corrosion
detection—which motivates our study of this model. We also note that these
techniques may be extended in a straightforward fashion to include more general parabolic equations. For example, one could incorporate a spatially-varying
thermal conductivity, and conduct the preceding analysis within the framework
of appropriately-weighted Hilbert spaces, with similar results.
38
Uniqueness for a Boundary Identification Problem
While a uniqueness result holds, this inverse problem is most certainly illposed; the shape of the region will not be a continuous function of the measured
data in any reasonable norm. A next logical step would be to examine and
quantify the nature of the ill-posedness and identify the features of the boundary
which can be stably estimated from the Cauchy data. This should give insight
into useful and practical reconstruction algorithms. Such an algorithm might
be based on the ideas in [2]—linearize the forward problem and examine the
linearized map from the “boundary shape” space to the measured temperature
data. The forward map will be given as an integral operator with smooth kernel,
and will have an unbounded inverse. We are currently investigating such an
approach to gain an understanding of stability and reconstruction possibilities.
References
[1] R.A. Adams, Sobolev Spaces, Academic Press, New York, 1975.
[2] K. Bryan and L.F. Caudill, Jr., An inverse problem in thermal imaging,
SIAM J. Appl. Math., 56 (1996), pp. 715–735.
[3] R. Cole, P. Shakespeare, H. Chissell, and S. Jones, Thermographic assessment of burns using a nonpermeable membrane as wound covering, Burns
17 (1991), pp. 117-122.
[4] D. Gilbarg and N. Trudinger, Elliptic Partial Differential Equations of Second Order (Second Edition), Springer-Verlag, Berlin, 1983.
[5] Y. Itoh and K. Arai, Use of recovery-enhanced thermography to localize
cutaneous perforators, Annals of Plastic Surgery 34 (1995), pp. 507-511.
[6] O. Ladyzhenskaya, V. Solonnikov, and N. Uraltseva, Linear and Quasilinear
Equations of Parabolic Type (Translations of Mathematical Monographs, v.
23), American Mathematical Society, Providence, RI, 1968.
[7] L. Nirenberg, Uniqueness in Cauchy problems for differential equations with
constant leading coefficients, Comm. Pure Appl. Math. 10 (1957), pp. 89105.
[8] J.C. Saut and B. Scheurer, Unique continuation for some evolution equations, J. Differential Equations 66 (1987), 118-139.
[9] I. Stakgold, Green’s Functions and Boundary Value Problems, John Wiley
& Sons, New York, 1979.
Kurt Bryan & Lester F. Caudill, Jr.
Kurt Bryan
Department of Mathematics
Rose-Hulman Institute of Technology
Terre Haute, IN 47803, USA.
E-mail: kurt.bryan@rose-hulman.edu
Lester F. Caudill, Jr.
Department of Mathematics and Computer Science
University of Richmond
Richmond, VA 23173, USA
E-mail: lcaudill@richmond.edu
39